/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Students may choose between a 3-... [FREE SOLUTION] | 91Ó°ÊÓ

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Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average examination grade of 84 with a standard deviation of \(4,\) and 18 students in the section without labs made an average: grade of 77 with a standard deviation of \(6,\) find a \(99 \%\). confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances.

Short Answer

Expert verified
The 99% confidence interval for the difference between the average grades for the two courses can be computed following these steps.

Step by step solution

01

Determine sample sizes, means, and standard deviations

First, recognize the given values, the sample sizes \(n_1\) and \(n_2\), the means \(\bar{x}_1\) and \(\bar{x}_2\), and the standard deviations \(s_1\) and \(s_2\) for labs and no labs classes respectively. Here, \(n_1 = 12\), \(\bar{x}_1 = 84\), \(s_1 = 4\), \(n_2 = 18\), \(\bar{x}_2 = 77\), \(s_2 = 6\).
02

Find the standard error of the difference

The standard error for the difference of two means can be found using the formula: \[SE = \sqrt{ \frac{{s_1}^{2}}{n_1} + \frac{{s_2}^{2}}{n_2} }\] Put the known values into the formula: \[SE = \sqrt{ \frac{{4}^{2}}{12} + \frac{{6}^{2}}{18} }\]
03

Find the 99% confidence interval

For a 99% confidence level, the z-score (z) is approximately 2.576 (usually found in statistical tables). Use the formula for the confidence interval for the difference of two means: \[CI = ((\bar{x}_1 - \bar{x}_2) - z * SE, (\bar{x}_1 - \bar{x}_2) + z * SE)\] Put the calculated standard error and the known values into the formula: \[CI = ((84 - 77) - 2.576 * SE, (84 - 77) + 2.576 * SE)\] to get the desired confidence interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error
The standard error (SE) is crucial in statistics as it measures the precision with which a sample mean represents the population mean. Essentially, it gauges the variation of the sample means around the true population mean. The lower the standard error, the more accurate the estimate.

In the context of comparing two means, the SE is modified to account for the variability within both samples. It's calculated using the formula: \[ SE = \sqrt{ \frac{{s_1}^{2}}{n_1} + \frac{{s_2}^{2}}{n_2} } \]where \(s_1\) and \(s_2\) are the standard deviations of the sample means, and \(n_1\) and \(n_2\) are the sample sizes. Knowing that the SE quantifies how much the difference between the sample means might vary, allows us to construct a range (interval) within which the actual difference between the population means likely falls.

For our exercise, calculating the SE is a step toward establishing how confident we can be about the difference in exam grades between students taking physics with labs versus without labs.
Difference of Two Means
The difference of two means is a statistical measure used to compare the averages from two distinct groups or samples. In the exercise, we're contrasting the average final examination grades for two groups of students: those who enrolled in a physics course with labs and those without.

When comparing two sample means, the actual point of interest is not the means themselves, but the difference between them. We denote this as \((\bar{x}_1 - \bar{x}_2)\). This difference has its own variability and distribution, which is where the standard error comes into play, providing a basis for how much this difference can fluctuate due to sampling error.

In the provided exercise, after identifying the sample means and their standard deviations, we then consider this difference under the assumption of normality, which allows us to employ the normal distribution's properties to construct a confidence interval around the difference.
Normal Distribution
A cornerstone of statistics is the normal distribution, a bell-shaped curve that is symmetric about the mean, depicting how values in a dataset are dispersed. Most values cluster around the mean, and as one moves away from the mean, the frequency of values decreases. It has many wonderful properties, one of which is that it's completely defined by two parameters: the mean and the standard deviation.

When considering the confidence interval for the difference of two means, we inexorably rely on the assumption that the distribution of these differences approximates, or is, normal. This assumption allows for the use of z-scores, which are standardized scores representing the number of standard deviations away from the mean a value lies.

Applying this to our exercise, we use the z-score corresponding to a 99% confidence level (approximately 2.576 for a two-tailed test) to indicate how far the difference of the sample means (our point estimate) should be adjusted to create an interval likely containing the true mean difference, given the normal distribution.

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Most popular questions from this chapter

A machine is used to fill boxes of product in an assembly line operation. Much concern centers around the variability in the number of ounces of product in the box. The standard deviation in weight of product is known to be 0.3 ounces. An improvement is implemented after which a random sample of 20 boxes are selected and the sample variance is found to be 0.045 ounces. Find a \(95 \%\) confidence interval on the variance in the weight of the product. Does it appear from the range of the confidence interval that the improvement of the process enhanced quality as far as variability is concerned? Assume normality on the distribution of weight of product.

Suppose that there are \(n\) trials \(x_{1}, x_{2}, x_{n}\) from a Bernoulli process with parameter \(p\), the probability of a success. That is, the probability of \(r\) successes is given by \(\left(\begin{array}{l}n \\\ r\end{array}\right) p^{r}(1-p)^{n-F} .\) Work out the maximum likelihood estimator for the parameter \(p\).

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a \(96 \%\) confidence interval for the population mean of all bulbs produced by this firm.

A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9 \(2.4,3.0,3.5,\) and 4.2 years, construct a \(95 \%\) confidence interval for \(\sigma^{2}\) and decide if the manufacturer's claim that \(\sigma^{2}=1\) is valid. Assume the population of battery lives to be approximately normally distributed.

A random sample of 10 chocolate energy bars of a certain brand has, on average, 230 calories with a standard deviation of 15 calories. Construct a \(99 \%\) confidence interval for the true mean calorie content of this brand of energy bar. Assume that the distribution of the calories is approximately normal.

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